Molality (in $m o l / k g$ ) of a solution in which 0.13 mole of solute is dissolved in $500 g$ of water is .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.26

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 0.26
We know,
Molality (m) $= \frac{N o o f g r a m m o l e s o f s o l u t e}{M a s s o f s o l v e n t (k g)}$
Mass of solvent is given to be $500 g (w a t e r)$
Number of moles of solute is given as 0.13.
$m = \frac{(0.13 \times 1000)}{500} = 0.13 \times 2 = 0.26$ .

Suggest Corrections

Similar questions

Molality (in mol/kg) of a solution in which 0.13 mole of solute is dissolved in 500 gm of water is:

m o l / k g

500 g

Join BYJU'S Learning Program